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+clc;
+P=2400;//N, Vertical Force applied at D
+AB=2.7;//m, perpendicular distance between A and B
+BE=2.7;//m, perpendicular distance between E and B
+BK=1.5;//m, perpendicular distance between B and K
+AJ=1.2;//m, perpendicular distance between A and J
+EF=4.8;//m, perpendicular distance between E and F
+BD=3.6;//m, perpendicular distance between D and B
+//For entire truss
+//By free body diagram we get the force at A, B , c
+A=1800;//N
+B=1200;//N
+C=3600;//N
+alpha=atan(EF/(AB+BE));//rad
+//a. Internal forces at j
+//Applying sum(M_J)=0
+M=A*AJ;//N.m,Couple on member ACF at J
+//Applying sum(Fx)=0
+F=A*cos(alpha);//N, Axial force at J
+//Applying sum(Fy)=0
+V=A*sin(alpha);//N, shearing force at J
+printf("Thus, Internal forces at J are equivalent to \n Couple M = %.0f N.m \n Axial force F= %.0f N \n Shearing force V= %.0f N\n",M,F,V);
+
+//a. Internal forces at K
+//Applying sum(M_K)=0
+M=B*BK;//N.m,Couple on frame
+//Applying sum(Fx)=0
+F=0;//N, Axial force at J
+//Applying sum(Fy)=0
+V=-B;//N, shearing force at J
+printf("Thus, Internal forces at K are equivalent to \n Couple M = %.0f N.m \n Axial force F= %.0f N \n Shearing force V= %.0f N\n",M,F,V);